Báo cáo toán hoc:"On the unitary Cayley graph of a finite ring" pot

On the unitary Cayley graph of a finite ringReza Akhtar Department of Mathematics Miami University reza@calico.mth.muohio.edu Megan Boggess Columbia Union College meganboggess@gmail.com

On the unitary Cayley graph of a finite ring

Reza Akhtar

Department of Mathematics Miami University reza@calico.mth.muohio.edu

Megan Boggess

Columbia Union College meganboggess@gmail.com

Tiffany Jackson-Henderson

St Augustine’s College tjackhend@yahoo.com

Isidora Jim´enez

Mills College ijimenez@mills.edu

Rachel Karpman

Scripps College rkarpman@scrippscollege.edu

Amanda Kinzel

Department of Mathematics Purdue University asphilli@math.purdue.edu

Dan Pritikin

Department of Mathematics Miami University pritikd@muohio.edu Submitted: May 16, 2009; Accepted: Sep 7, 2009; Published: Sep 18, 2009

Mathematics Subject Classification: 05C25, 05C30

Abstract

We study the unitary Cayley graph associated to an arbitrary finite ring, de-termining precisely its diameter, girth, eigenvalues, vertex and edge connectivity, and vertex and edge chromatic number We also compute its automorphism group, settling a question of Klotz and Sander In addition, we classify all planar graphs and perfect graphs within this class

1 Introduction

Given an integer n, consider the graph Cay(Zn, Z∗

n) with vertex set Zn (the integers modulo n), with vertices x and y adjacent exactly when x − y is a unit in (the ring) Zn These so-called unitary Cayley graphs have been studied as objects of independent interest (see, for example, [2], [3], [7], [8], [9]) but are of particular relevance in the study of graph

representations, begun in [5] and continued in many other papers A graph is said to be representable modulo n if it is isomorphic to an induced subgraph of Cay(Zn, Z∗

n); the central problem in graph representations is to determine the smallest positive n modulo which a given graph G is representable It is natural, then, to study unitary Cayley graphs

in the hope of gaining insight into the graph representation problem

A generalization of unitary Cayley graphs presents itself readily: given a finite ring R (commutative, with unit element 1 6= 0), one may define GR= Cay(R, R∗) to be the ring whose vertex set is R, with an edge between x and y if x − y ∈ R∗ This construction was introduced in [7] and [8], although it does not appear to have been considered in [9] This article began as a project to address the question of computing the automorphism group Aut(Cay(Zn, Z∗

n)), first raised by Klotz and Sander in [9] We soon realized that

it was more natural to consider this question in the context of (unitary Cayley graphs of) finite rings In this article we give a complete answer to this question; moreover,

we extend the results of [9] to the setting of finite rings and explore various other graph-theoretic properties not considered there Our proofs emphasize the dependence of results

on the underlying algebraic structure of the rings concerned; in some cases, these provide

a considerable simplification of the Klotz-Sander proofs We hope that the use of some algebra will provide a more cohesive approach to further study of these graphs

A key observation is the following: since R is a finite ring, it is Artinian, and hence

R ∼= R1 × × Rt, where each Ri is a finite local ring with maximal ideal mi Since (u1, , ut) is a unit of R if and only if each ui is a unit in R∗

i, we see immediately that GR is the conjunction (sometimes called tensor product or Kronecker product) of the graphs GR 1, , GR t Moreover, if x, y ∈ Ri, {x, y} is an edge of GR i if and only if

x − y 6∈ mi It follows immediately that each GRi is a complete balanced multipartite graph whose partite sets are the cosets of mi in (the additive group) Ri This perspective allows us, for example, to give a simple, explicit computation of the eigenvalues of the graphs GR (see Section 10) In future work we hope to generalize the study of graph representations to this broader setting

Part of this research was carried out in the SUMSRI program, held at Miami University during the summer of 2008 We thank Miami University, the National Security Agency, and the National Science Foundation for their support of the first six authors We also thank the referee for suggestions which helped improve this paper

2 Algebraic Background and Basic Properties

Throughout this paper, all rings mentioned are commutative with unit element 1 6= 0 Let R be a finite ring Since R is Artinian, the structure theorem [4, p 752, Theorem 3] implies that R ∼= R1 × × Rt, where each Ri is a finite local ring with maximal ideal mi; this decomposition is unique up to permutation of factors We denote by ki the (finite) residue field Ri/mi, πi : Ri → ki the quotient map, and fi = |ki| We also assume (after appropriate permutation of factors) that f1 6f2 6 6 ft This notation will be

maintained throughout the paper whenever R is mentioned as a finite (or more generally Artinian) ring

The following proposition is well-known, but we include it here for the sake of com-pleteness

Proposition 2.1 Let S be a finite local ring with maximal ideal m Then there exists a prime p such that |R|, |m| and |R/m| are all powers of p

Proof

Since k = R/m is a field, its order must be equal to pe for some prime p and integer

e > 1 Since R is finite, Nakayama’s Lemma implies that as long as mi 6= 0, mmi =

mi+16= mi; that is, mi+1 is a strict subset of mi Since R is finite, this implies that in the chain R ⊇ m ⊇ m2 ⊇ , there is some i such that mi = 0 Then for all j > 1, |mj−1/mj|

is a k-vector space, so its order is a power of p Then descending induction on j shows that |mj| is a power of p for all j > 0

We note in particular that the nilradical of a local ring R (the ideal NR of nilpotent elements) is simply the (unique) maximal ideal of R

It is well-known that if R is an Artinian ring, then R ∼= R1 × × Rt, where each

Ri is an Artinian local ring Furthermore, R∗ = R∗

1 × × R∗

t, and hence two vertices

x = (x1, , xt), y = (y1, , yt) are adjacent if and only if xi− yi ∈ R∗

i for all i = 1, , t Equivalently, x is adjacent to y if and only if for each i = 1, , t, xi− yi 6∈ mi; that is,

πi(xi) 6= πi(yi)

The following are some basic consequences of this definition:

Proposition 2.2

• Let R be any ring Then GR is a regular graph of degree |R∗|

• Let S be a local ring with maximal ideal m Then GS is a complete multipartite graph whose partite sets are the cosets of m in S In particular, GS is a complete graph if and only if S is a field

• If R is any Artinian ring and R ∼= R1 × × Rt as a product of local rings, then

GR=Vt

i=1GR i Hence, GR is a conjunction of complete multipartite graphs Proof

The first statement follows from the fact that the neighborhood of any vertex a is {a + u : u ∈ R∗} For the second statement, simply note that x, y ∈ S are adjacent if and only if x − y 6∈ m and that S is a field if and only if m = 0 The third statement follows from the fact that R∗ ∼= R∗

1× × R∗

t

For any r ∈ R, the map z 7→ z + r defines an automorphism GR; similarly, if u ∈ R∗,

z 7→ uz is also an automorphism of GR We will compute the full group Aut(GR) in Section 4

Throughout this paper, we use N(v) for the neighborhood of a vertex (that is, the set

of vertices adjacent to v) and N(u, v) for the number of common neighbors of the vertices

u and v We now give a formula for the latter in GR:

Proposition 2.3 Suppose a = (a1, , at) and b = (b1, , bt) are vertices of GR Let

I = {i : 1 6 i 6 t, πi(ai) = πi(bi)} and J = {1, , t} − I Then

N(a, b) = |R|Y

i∈I

(1 − 1

fi

)Y

j∈J

(1 − 2

fj

)

Proof

If c = (c1, , ct) is adjacent to both a and b, then for each k = 1, , t, ci may be any element such that πi(ci) 6∈ {πi(ai), πi(bi)} If πi(ai) = πi(bi), there are fi− 1

fi

|Ri| choices

for ci, and if πi(ai) 6= πi(bi), there are fi− 2

fi

|Ri| choices for ci In total, then, there are Y

i∈I

(1 − 1

fi

)|Ri| ·Y

j∈J

(1 − 2

fj

)|Rj| = |R|Y

i∈I

(1 − 1

fi

)Y

j∈J

(1 − 2

fj

) choices for c

Corollary 2.4 Let R be an Artinian ring and x, y ∈ GR Then N(x) = N(y) if and only if x − y ∈ NR

Some questions about properties of unitary Cayley graphs are best viewed as purely combinatorial questions about conjunctions of complete balanced multipartite graphs We will adopt this persective at various points later in this article Sometimes we can simplify things even further, as explained in the next paragraph

Consider two vertices v, w of a graph G to be equivalent when N(v) = N(w) Then, following [6] we define the reduction of G to be the graph Gred whose vertex set is the set of equivalence classes of vertices (as defined above), and whose edges consist of pairs {A, B} of equivalence classes with the property that A ∪ B induces a complete bipartite subgraph of G

Proposition 2.5 Let R be an Artinian ring Then the reduction (GR)red ∼= GR red where

Rred = R/NR is the (ring-theoretic) reduction of R

Proof

First, write R = R1× .×Rtas a product of local rings Then NR= NR 1× .×NR t =

m1× × mt and so Rred = R1/m1 × × Rt/mt is a product of fields Moreover, it is clear from the description of adjacency above that two vertices (a1, , at), (b1, , bt) of

GRhave the same neighborhood if and only if ai− bi ∈ mi for all i = 1, , t This implies

that vertices of (GR)red correspond to elements of Rred = R1/m1 × × Rt/mt Since adjacency is defined by the same rule in both graphs, it follows that (GR)red∼= GR

red Proposition 2.5 allows us to convert general questions about unitary Cayley graphs of finite rings to corresponding questions about finite reduced rings (i.e products of fields)

3 Diameter and Girth

In the following we use diam(G) and gr(G) (respectively) to denote the diameter and girth of a graph G

Theorem 3.1 Let R = R1× × Rt be an Artinian ring Then

diam GR =















1 if t = 1 and R is a field

2 if t = 1 and R is not a field

2 if t > 2, f1 >3

3 if t > 2, f1 = 2, f2 >3

∞ if t > 2, f1 = f2 = 2

Proof

If t = 1, then by Proposition 2.2, GR is complete if R is a field, and is a complete multipartite graph (with at least two partite sets) if R is not a field In the first case, GR

has diameter 1; in the second case, it has diameter 2 Now suppose t > 2 and f1 > 2 Then

fi >3 for all i = 1, , t, so given distinct vertices a = (a1, , at), b = (b1, , bt), select elements ci ∈ Ri, i = 1, , t, such that πi(ci) 6∈ {πi(ai), πi(bi)} Then c = (c1, , ct) is a common neighbor of a and b and so diam GR 62 Obviously GR is not complete in this case, so diam GR = 2

If t > 2 and f1 = 2, observe that the vertices (0, 0, , 0) and (1, 0, , 0) are neither adjacent nor do they share a common neighbor; hence diam GR>3 If, moreover, f2 = 2, then there is a no path in GRbetween these same two vertices, so GRis disconnected On the other hand, if f2 > 3, consider distinct vertices a = (a1, , at) and b = (b1, , bt) such that d(a, b) > 3 In particular, π1(a1) 6= π1(b1) and for some i > 2, πi(ai) =

πi(bi) Now define c = (c1, , ct), d = (d1, , dt) as follows: for each i, 1 6 i 6 t, if

πi(ai) = πi(bi), pick ci, di ∈ Ri such that πi(ci), πi(di), and πi(ai) = πi(bi) are distinct;

if πi(ai) 6= πi(bi), set di = bi and ci = ai Then a, d, c, b is a path of length 3, so diam GR= 3

Theorem 3.2 gr GR=











3 if f1 >3

6 if R ∼= Zr

2× Z3 for some r > 1

∞ if R ∼= Zr2 for some r > 1

4 otherwise

Proof

Suppose first that f1 > 3 Then any three vertices a = (a1, , at), b = (b1, , bt),

c = (c1, , ct) such that πi(ai), πi(bi), and πi(ci) are distinct for all i = 1, , t induce a triangle, and so gr GR = 3

We next consider the case t = 1, f1 = 2 If R ∼= Z2, clearly gr GR= ∞ Otherwise, R

is not a field, so GR is a complete bipartite graph with partite sets of size |m1| > 2, and hence gr GR = 4

Now suppose f1 = 2 and t > 2 Then GR is a bipartite graph, so gr (GR) > 4 Let

a = (0, , 0) and b = (1, , 1) If Ri is not a field for some i > 1, then |mi| > 2,

so choosing 0 6= x ∈ mi, define c = (c1, , ct) and d = (d1, , dt) by setting, for each

j = 1, , t, cj = δijx and dj = 1 + δijx Then a, b, c, d, a is a 4-cycle, and so gr (GR) = 4

If Rj is a field for all j > 1 and |Ri| > 4 for some i, choose elements ci, di ∈ Ri such that

πi(ci), πi(di) are distinct elements of k − {0, 1} For j 6= i, define cj = aj and dj = bj, and let c = (c1, , ct), d = (d1, , dt) Then a, b, c, d, a is a 4-cycle, and so gr (GR) = 4 in this case, too

We are now reduced to the case that R ∼= Zr2× Zs

3 for some r, s, r + s > 2 Since GR

is bipartite, it contains no odd cycles To simplify notation in the following discussion

we use the notation xm to represent an m-tuple each of whose coordinates is x (in the appropriate ring) If s > 2, then (0r, 0s), (1r, 1s), (0r, 2, , 2, 0), (1r, 1, , 1, 2), (0r, 0s) defines a 4-cycle in GR If s = 1, the vertex sequence (0r, 0), (1r, 1), (0r, 2), (1r, 0)

(0r, 1), (1r, 2), (0r, 0) defines a 6-cycle, so gr GR 6 6 If a, b, c, d, a were a cycle of length

4 in GR, then ai = ci for all i and bi = di (1 6 i 6 r), so in particular ar+1 6= cr+1,

br+1 6= dr+1, and so S = {ar+1, cr+1} and T = {bd+1, dr+1} are (by virtue of the adjacency conditions) disjoint subsets of Rr+1, each of cardinality 2 However, |Rr+1| = 3, so this is

a contradiction Thus, gr GR= 6 The last case to consider is R ∼= Zr2, but in this case,

GR∼= 2r−1K2 and hence gr GR = ∞

Corollary 3.3 The number of triangles in GR is |R|

3

6

t

Y

i=1

(1 − 1

fi

)(1 − 2

fi

)

Proof

If f1 = 2, then by Proposition 3.2, GR is triangle-free, so the claim holds in this case

If f1 > 3, then given a vertex a ∈ R, by Proposition 2.2 there are |R∗| = |R|

t

Y

i=1

(1 − 1

fi

) choices for an adjacent vertex b Now, Proposition 2.3 implies that there are |R|

t

Y

i=1

(1−2

fi

) choices for a third vertex which is a common neighbor of both a and b Since any such triangle may be formed in 6 distinct ways, the total number of triangles is |R|

3

6

t

Y

i=1

(1 − 1

fi)(1 −

2

fi).

4 Automorphisms

In this section we compute the group Aut(GR) when R is a finite ring We begin by reducing the problem to the case of reduced rings

Lemma 4.1 Let R be a finite ring and n = |NR|.

Then there is an isomorphism f : Aut(GR)→ Aut(G∼ R red) × (Sn)|R/N R |

Proof

It follows from Corollary 2.4 that any σ ∈ Aut(GR) permutes the cosets of NR in R; in particular, σ induces an automorphism ¯σ ∈ Aut(GR red) Moreover, if one fixes

an enumeration x1, , xn of the elements of NR and a set of coset representatives R = {aC}C∈R/NR, then for each such coset C = aC + NR of NR in R, σ(C) = bC + NR for some representative bC ∈ R; in particular, there is a permutation σC ∈ Sn such that for each i = 1, , n, σ(aC + xi) = bC + xσ C (i) We now define f (σ) = (¯σ,Q

C∈R/N RσC); it is immediate that f is a homomorphism and that Ker f = 1, so f is injective

Now suppose we are given ψ = (τ,Q

C∈R/N RφC) ∈ Aut(GR red) ×Q

C∈R/N RSn By construction, each element of R may be written uniquely as aC+ xj for some C ∈ R/NR

and 1 6 j 6 n Define bC to be the (unique) element of R satisfying τ (aC+NR) = bC+NR Now define σ ∈ Aut(GR) by σ(aC + xj) = bC + φC(xj) Then f (σ) = ψ and so f is surjective

For rings S1, , Sm, we define the number of leading zeros of an element s = (s1, ,

sm) ∈ S1× × Sm to be max{ℓ > 0 : s1 = = sℓ = 0}

We now turn to the case of reduced rings

Theorem 4.2 Let s > 1, and suppose r1, , rs are prime powers such that 2 6 r1 < < rs For each i = 1, , s, let ni > 1 be an integer, and consider the ring R =

Qs

i=1(Fi)n i, where Fi denotes the field with ri elements Then Aut(GR) ∼= Qs

i=1Sri ×

Qs

i=1Sn i

Proof

The idea behind the proof is to identify certain “obvious” automorphisms of GR and then prove that any automorphism σ coincides with one of these, using the property that for any two vertices u, v ∈ GR, N(σ(u), σ(v)) = N(u, v)

Since R is reduced, any (set) map f : R → R which fixes all but one of the local factors and permutes the elements of the remaining factor induces an automorphism of

GR Similarly, a map f : R → R which is the identity on (Fi)n i for i 6= i0 and permutes the ni 0 factors of the form Fi 0 induces an automorphism of GR Let H ⊆ Aut(GR) be the subgroup generated by maps of either of these two types It remains to check that in fact

H = Aut(GR) Observe that translations, i.e automorphisms of the form z 7→ z + a for some fixed a ∈ R, are compositions of maps of the first type

To this end, suppose σ ∈ Aut(GR) Composing with a translation, we may assume without loss of generality that σ(0) = 0 Our goal is prove that, after composition with maps in H, σ(a) = a for all a ∈ R We do this by downward induction on the number ℓ

of leading zeros in the coordinate representation for a, the base case being ℓ = m; that is,

a = 0

Suppose by induction that σ(a) = a for all a with more than ℓ leading zeros, and sup-pose b = (b1,1, , b1,n 1, , bs,1, , bs,n s) ∈ R has ℓ leading zeros Suppose the leftmost

nonzero coordinate in b is the (i, j) coordinate Define b′ to have the same coordinates

as b except for the (i, j) coordinate, which is 0 Observe that if c ∈ R has at most ℓ leading zeros, then |N(b, b′)| 6 |N(c, b′)| by Proposition 2.3, with equality if and only

if c and b differ only in the (i, k) coordinate for some k, 1 6 k 6 j By induction, σ(b′) = b′ and σ(b) has at most ℓ leading zeros Moreover, since σ is an automorphism, N(b, b′) = N(σ(b), b′), so by the inequality above, σ(b) differs from b only in the (i, k) coordinate, where 1 6 k 6 j By applying an automorphism in H of the second type, we may assume that k = j, and after applying an automorphism of the first type, σ(b) = b This completes the induction

5 Connectivity

Proposition 5.1 Let R be any finite ring, and let κ(GR) and κ′(GR) denote (respectively) the vertex-connectivity and edge-connectivity of its unitary Cayley graph Then κ(GR) =

κ′(GR) = |R∗|

Proof

We argue following the reasoning in [9], Theorem 4 According to a theorem of Watkins [10], the vertex connectivity of a regular edge-transitive graph is equal to its degree of regularity We show that GR is edge-transitive by observing that for any edge {u, v} the automorphism x 7→ (v − u)−1(x − u) maps u to 0 and v to 1 Hence κ(GR) = |R∗| Since κ(GR) 6 κ′(GR) 6 |R∗| by [11, Theorem 4.1.9], it follows that κ(GR) = κ′(GR) = |R∗|

6 Clique Number, Chromatic Number, and Indepen-dence Number

For a graph G, we denote by ¯G its complement, ω(G) its clique number, α(G) its inde-pendence number, and χ(G) its chromatic number

Proposition 6.1 Let R be a finite ring Then ω(GR) = χ(GR) = f1 and ω(GR) = χ(GR) = α(GR) = |R|

f1

Proof

Choose elements rij ∈ Ri, i = 1, , t, j = 1, , f1 such that for each i = 1, , t and j 6= j′, πi(rij) 6= πi(rij′) Then, setting aj = (r1j, , rtj) for each j = 1, , f1, it

is easily seen that C = {a1, , af 1} is a clique and ω(GR) > f1 Now consider the ideal

I = m1×R2 .×Rt⊆ R There are precisely f1 cosets of I in R, each of which corresponds

to an independent subset of GR By assigning each coset a distinct color and coloring all vertices within that coset the same color, we have constructed a proper coloring of GR Hence, χ(GR) 6 f1 Since f1 6ω(GR) 6 χ(GR) 6 f1, we have ω(GR) = χ(GR) = f1

Since the ideal I constructed above corresponds to an independent set in GR, we have α(GR) = ω(GR) > |I| = |R|/f1 We now construct a coloring of GR by elements of I as follows: given b = (b1, , bn) ∈ R, fix a clique C in GRas above and let cb be the unique element of C such that b − cb ∈ I; define a vertex coloring f : R → I by f (b) = b − cb Then f (b) = f (d) implies that b − d = cd− cb If cd= cb, then b = d; so assume cd 6= cb Then by construction, cd− cb ∈ R∗, so b − d ∈ R∗, and hence b is not adjacent to d in GR Thus f is a proper coloring, showing that χ(GR) 6 |I| = |R|/f1, as desired

Corollary 6.2 Let R be a finite ring Then GR is f1-partite

7 Edge Chromatic Number

We next derive a result concerning the edge chromatic number χ′(GR)

Theorem 7.1 Let R be a finite ring Then

χ′(GR) =



|R∗| + 1 if |R| is odd

|R∗| if |R| is even

Proof

Since GR is |R∗|-regular, χ′(GR) > |R∗|, and by Vizing’s Theorem, χ′(GR) 6 |R∗| + 1 Suppose |R| is odd, so GR has no 1-factor Then in any proper edge-coloring of GR, each color class must miss some vertex x Hence there are |R∗| colors used on edges incident

at x, plus the color of that class used elsewhere; hence, χ′(GR) = |R∗| + 1

Now suppose |R| is even By Proposition 2.1, at least one of the local rings in the decomposition R ∼= R1× × Rt has even cardinality In particular, this means that for any unit u = (u1, , ut) ∈ R∗, |u| = lcm(|u1|, , |ut|) is even, where by |u| (or |ui|) we mean the order of u as an element of the additive abelian group R (respectively, Ri) Let V = {v ∈ R∗ : |v| = 2} and EV = {{r, r + v} : v ∈ V } ⊆ E(GR) We observe that there are exactly |V | edges of EV incident at every vertex of GR Now construct

a proper coloring of E(G) as follows: fix a bijection h : V → {1, , |V |} and, for each

v ∈ V and r ∈ R, color the edge {r, r + v} with color h(v) Now let U′ = R∗− V ; note that for each u ∈ U′, u 6= −u Choose U = {u1, , um} ⊆ U′ such that for all u ∈ U′, exactly one of u, −u is in U; thus, |U| = |R

∗| − |V |

2 Now for each uj, j = 1, , m, let a1+ < uj >, , as+ < uj > be the (distinct) cosets of < uj > in R For each

k = 0, , |uj| − 1, color the edge {ai+ kuj, ai+ (k + 1)uj} with color |V | + 2j − 1 if k is odd or color |V | + 2j if k is even It is easy to check that this procedure defines a proper edge-coloring of G with 2|U| + |V | = |R∗| colors

8 Planarity

The following is immediate from definitions:

Lemma 8.1 Let G be a bipartite graph Then G ∧ K2 ∼= 2G In particular, G is planar

if and only if G ∧ K2 is planar

Our result on planarity is:

Theorem 8.2 Let R be a finite ring Then GR is planar if and only if R is one of the following rings: (Z/2Z)s, Z/3Z × (Z/2Z)s, Z/4Z × (Z/2Z)s, or F4× (Z/2Z)s (Here F4

is the field with four elements and s > 0 may assume any integer value.)

Proof

Clearly, Z/2Z and Z/3Z are the only rings with fewer than 4 elements, so henceforth let R be a finite ring such that GR is planar and |R| > 4

If f1 = 2, then R ∼= R1× × Rt is bipartite by Corollary 6.2 and as such is triangle-free By a well-known result (see for example [11, Theorem 6.1.23]), planarity of R forces

|E(R)| 6 2|R| − 4; that is, |R∗| 6 4 − 8

|R| or |R

∗| 6 3 Now if S is a local ring, |S∗| > |S|

2 ; hence, |S∗| = 1 if and only if S ∼= Z/2Z Moreover, R∗ = R∗

1× × R∗

t, so the condition

|R∗| 6 3 forces R ∼= S × (Z2)s for some s > 0 and some local ring S with |S∗| 6 3 Since

|S| 6 6 and |S| must be a prime power, the only possibilities are S = Z/2Z, Z/3Z, Z/4Z (with any s > 0), or S = F4 (with any s > 1) It is easy to check by hand that for each of these choices of S, both GS and GS×Z/2Z are planar Since the graph GS×Z/2Z is guaranteed to be bipartite by Corollary 6.2, planarity of S × (Z2)s follows from Lemma 8.1 by induction

Now suppose f1 >3 Then (cf [11, Theorem 6.1.23]) planarity of R ∼= R1× × Rt forces |E(R)| 6 3|R| − 6, which implies |R∗| 6 5 However, this time each of the local factors Ri satisfies |R∗i| > 2

3|Ri|; in particular, if |R

∗

i| = 2, then |Ri| = 3 and hence

Ri ∼= Z/3Z, which is impossible since f1 > 3 If |R∗| = 3, then R ∼= F4, and if |R∗| = 4, then R ∼= Z/5Z Clearly GF 4 ∼= K4 is planar but G

Z/5Z ∼= K5 is not.

9 Perfectness

Let R be an Artinian ring In this section, we classify which of the graphs GRare perfect

As before, fix a decomposition R ∼= R1× × Rtas a product of local rings We note that our proof, while following the outline of the analogous result in Section 3 of [9], differs somewhat in that it avoids use of the Fuchs-Sinz result [8] on longest induced cycles

If t = 1 then by Proposition 2.2 GR is complete multipartite and hence is perfect If

f1 = 2 then by Corollary 6.2 GR is bipartite and hence perfect We assume henceforth that f1 >3 Our main tool is the Strong Perfect Graph Theorem

Theorem 9.1 [3] A graph G is perfect if and only if neither G nor ¯G contains an induced odd cycle

Lemma 9.2 Suppose t > 3 Then GR is not perfect